07 Ideal Bose a
07.01

The septic heats of solids : the phonon gas: nd Fermi Systems
The energy of the elastic waves or sound waves inside a solid medium can be considered to be
quantized in the form of phonons. The energy of phonons of frequency υ is hυ and because the phonons
have integral angular momentum. the assembly of phonons in the solid may be treated as a boson gas.
The general from of Bose Einstein distribution function is
1
nj =
e
 ( j   )
1
Where n j is the average occupation number in the state j with energy
 j; β =
1
, T being the temperature of
kT
the system in equilibrium ; μ is the chemical potential of the system. For a Phonon gas in a solid the total
number of phonons is not necessarily conserved because an excited state of a normal mode ( represented by a
harmonic oscillator ) may contain any number of quanta. Since the total energy of the solid remains constant.
We must take μ =o for a phonon gas. Thus we have for phonon gas.
nj =
1
e
 i | kT
1
If a solid has N atoms, it has 3N normal modes of oscillations. Therefore there
will be 3N different types of phonon with characteristic frequencies
1 , 2 , 3 , ........................... 3 N The values of these frequencies depend on the nature of the lattice. In the
Einstein model these frequencies are taken to be equal to one another
1 = 2 ......... = 3 N
= .
That is only one state of energy   h and degeneracy 3N is available for a phonon. Dropping the subscript j
in equation (2) we then get that the total energy of the photon gas is
U=(3N)(h  )
1
e
h | kT
1
...............................(3)
Heat Capacity at constant volume will then be
Cv
 u 


 T v
 3Nh
e
1
h | kT

1
2
e
h
kT
 h 
 2 
 kT 
e h |kT
 h 
3NK 
 h |kT
2
 kT  e
1
2
Cv =


This is the required expression.
for one mole of a solid N=N0 = 6.023 1023 . So that have the molar specific heat formula
e h | kT
 h 
Cv = 3R 

2 ;
 kT  e h | kT  1
2
R
 N.K
07.02

The Fermi-Drac Gas
Consider an ideal gas of Fermions with spin’s enclosed in a volume V at equilibrium temperature T= 0
k. At absolute zero, the Fermi function f (t) giving the average occupation number of fermions that
have energy
, d  becomes.
f () =  F 0  t  
0  F 0 
1
 F 0 
............... (1)

Where F 0 1s the Fermi energy at absolute zero. Here we have assumed a continum of energy
states. The number of states a for which Fermions momentum lies in the interval (p, p +dp) is
given by

g(p)dp =
4p dp 
2
h3
This expression must be multiplied by a factor (2s+1), since fermions of the same momentum can have
(2s+1) different orientations. Then
g(p) dp = (2s+1)

4p dp  ............................. (2)
2
h3
But energy  corresponding to momentum p is
=
p2
2m
or
p  2m 
Where m is the mass of each fermions There fore the total number of states with energy value in the
range , d  is


g d  2 S  1
Now if

h
3
2m 3 / 2 2
 d  ............................... (3)
N is the total number of fermions in the system we must have,

N   f ()g ()d 
0

=
2S  12 2m3 / 2 3 
h
 (F 0   )
0
F  0 
=
2s  12 (2m)3 2 3 
h
d 
0
 



h 3
4
2m    0 
N = (2s+1)
3
h
= 2 s  12 ( 2m)
32
 2
3
0
32
3
or,
F 0
2
F
32
32
3
F
3h3 N
=
2s  14 (2m)3 2 .v


3h3 N
or, F 0 = 
32 
 4 (2s  1)( 2m) .v 
h 2  6( N / v) 
 F 0=


8m   (2s  1) 
23
23
This is the Fermi energy for an ideal Fermi gas at absolute zero.
In Particular if we have an electron gas then S=
 
= ђ /2m 3 N / v 
F (o) 
h2  3

N /v 

8m  

2
2
23
23
1
thus
2